Question

prove that cos^3x+3cos2x= 4 (cos^6 x-sin^6x)

Answer 1

But this isn't true... If $x=\dfrac\pi2$, you have

$\cos^3\dfrac\pi2+3\cos\pi=0-3=-3$
$4\left(\cos^6\dfrac\pi2-\sin^6\dfrac\pi2\right)=4(0-1)=-4$

but of course $-3\neq-4$.